In: Chemistry
1. It is estimated that codeine has a maximum water solubility
of 9000 mg/L.
a) Calculate the pH and %
ionization of a saturated solution of codeine (pKb =
5.80).
b) Will more, or less, codeine dissolve when adding it to a
solution containing…
i. a strong acid? Explain!
ii. a strong base? Explain!
Assume max solubility
S = 9000 mg/L = 9 g /L
MW of cdeine = 299.364 g/mol
[Molar] = S/MW = (9g/L) / (299.364 g/mol) = 0.03 M
then...
a)
pH in saturation
pKb = 5.80, therefore a base...
This is a base in water so, let the base be codeine= "B" and codeineH+ = HB+ the protonated base "HB+"
there are free OH- ions so, expect a basic pH
B + H2O <-> HB+ + OH-
The equilibrium Kb:
Kb = [HB+][OH-]/[B]
initially:
[HB+] = 0
[OH-] = 0
[B] = M
the change
[HB+] = x
[OH-] = x
[B] = - x
in equilibrium
[HB+] = 0 + x
[OH-] = 0 + x
[B] = M - x
Now substitute in Kb
Kb = [HB+][OH-]/[B]
Kb = x*x/(M-x)
x^2 + Kbx - M*Kb = 0
x^2 + (10^-5.80)x - (0.03)(10^-5.80) = 0
solve for x
x = 2.17*10^-4
substitute:
[HB+] = 0 + x = 2.17*10^-4 M
[OH-] = 0 + x = 2.17*10^-4 M
[B] = M - x = 0.03- 2.17*10^-4 = 0.029783 M
pH = 14 + pOH = 14 + log( 2.17*10^-4 ) = 10.34
for ionization
% ionizaitn of base = [HB+] / M * !00 %= (2.17*10^-4) /(0.03) * 100 = 0.7233 %
b)
if we add more storng acid:
there will be neutralization of the base
the solubility equilibrium
Codeine(s) --> Codeine(aq) will be ffaovured towads the RIGHT, since codeine is being neutralized, so MORE solid must go into solution
for the strong base:
This will be backwards,
the equilibrium:
Codeine(aq) + OH- --> Codeine(s) is favoured
The OH- sautrates so forms more codeine in aqueus solution
so the solubility must decrease